ANOVA Table
Source of variation |
Sum of Squares (SS) |
Degrees of freedom |
Mean Square (MS) |
F |
Treatment |
93.3 |
2.0 |
46.67 |
8.75 |
Error |
64.0 |
12.0 |
5.33 |
|
Total |
157.3 |
14 |
SAT scores - 2017
Student |
Reading |
Math |
Writing |
1 |
453 |
458 |
465 |
2 |
456 |
458 |
459 |
3 |
454 |
460 |
457 |
4 |
458 |
455 |
461 |
5 |
454 |
454 |
463 |
mean |
455 |
457 |
461 |
Solution:
Given that , = 455,
= 457
= 461
Alpha = 0.05,
df error = 12 and MS error = 5.33
na = nb = 5
The value for Alpha = 0.05, df = 12 is 2.1788
LSD = t critical sqrt(MSE (1/na + 1*/nb) = 2.1788 sqrt[5.33 (1/5
+ 1/5)]
LSD = 3.18
If the absolute value difference of Means (taking 2 at a time ABS(Mi - Mj) is \geq LSD, then reject the hypothesis that the means are the same.
No, we cannot agree with the results that the mean are the same, because after conducting Fishers LSD, we see that there is a significant difference between Mean 1 and Mean3 and between Mean2 and Mean 3.
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